# stuck with equivalent resistance

Thread Starter

#### clangray

Joined Nov 4, 2018
261
This is super basic. Thanks in advance.

Can the two resistors in the diagram have equivalent resistance or do you find a third value from those two that is the equivalent resistance?

#### BobTPH

Joined Jun 5, 2013
8,104
They are neither in series nor parallel, so I don’t know what you mean by equivalent resistance here.

If you you mean “what resistor would you need to get the equivalent result if the two strings if LEDs were in parallel,” the answer is 83 Ohms.

#### Irving

Joined Jan 30, 2016
3,548
But you'd never put the LED strings in parallel...

Maybe the TS meant the equivalent resistance of the whole circuit? ie 5v / 30.5mA = 163.9Ω It makes as much sense as anything...

#### WBahn

Joined Mar 31, 2012
29,506
This is super basic. Thanks in advance.

Can the two resistors in the diagram have equivalent resistance or do you find a third value from those two that is the equivalent resistance?
The notion of "equivalent resistance", at its essence, means to be able to replace some set of resistors with a single resistor such that the rest of the circuit can't tell the difference. Imagine making a copy of the circuit and putting part of it in a box (or under a paper tent, or whatever else you might use to hide part of it). Now imagine that someone comes in and replaces what is in the box on one of the circuits with something else. You are then tasked with coming in and determining which circuit was changed, but you can't look in the box in any way. All of your work, be it measurements or modifications, have to be to the part of the circuit you can see. If it is impossible for you two tell the difference, then what is in the box on the modified circuit is truly equivalent to what is in the box on the original circuit.

If the portion of the circuit that is in the box is linear, then coming up with a true equivalent is pretty straightforward, but if it is nonlinear, it becomes trickier. But sometimes we don't want/need something that is truly equivalent, but rather something that is equivalent (or approximately so) only under certain conditions.

You have a nonlinear circuit, due to the presence of the diodes. So it becomes particularly important to identify what portion of the circuit you are looking for something that is equivalent and under what conditions you care about. The best place to start on that is to draw a box around the portion of the circuit you want to replace with an equivalent. That will help us focus the discussion.

#### dl324

Joined Mar 30, 2015
16,143
Can the two resistors in the diagram have equivalent resistance
The resistors can't be combined.

If you mean can you use a single resistor and have the two strings of LEDs connected in parallel; that's not advisable.

Manufacturers do it to save money at the expense of having an unreliable product. I have a number of flashlights with parallel LEDs and some of them only have a few LEDs that are providing useful light.

#### DickCappels

Joined Aug 21, 2008
10,104
You can combine the two strings IF and only if the components are essentially identical. This rarely happens in real life.

Thread Starter

#### clangray

Joined Nov 4, 2018
261
They are neither in series nor parallel, so I don’t know what you mean by equivalent resistance here.
I thought they were parallels at best. But its neither? How do I make them parallel then?
If you mean can you use a single resistor and have the two strings of LEDs connected in parallel; that's not advisable.
I want 2 strings of LEDs in parallel and a power LED power on/off(not shown), not a single resistor to 2 parallels.
The reason behind selecting the parallel string is due to its orientation. I would use series if I could.

Thread Starter

#### clangray

Joined Nov 4, 2018
261
Maybe the TS meant the equivalent resistance of the whole circuit? ie 5v / 30.5mA = 163.9Ω It makes as much sense as anything...
But you'd never put the LED strings in parallel...

Maybe the TS meant the equivalent resistance of the whole circuit? ie 5v / 30.5mA = 163.9Ω It makes as much sense as anything...
Yes, I was trying to solve the circuit resistance in parallel based on the circuit 5V and current.

#### WBahn

Joined Mar 31, 2012
29,506
I thought they were parallels at best. But its neither? How do I make them parallel then?
In order to be in parallel, they have to be connected such that the same voltage that appears across one must appear across the other. Not the same value of voltage, but the actual same voltage (sometimes referred to as the same symbolic voltage).

I want 2 strings of LEDs in parallel and a power LED power on/off(not shown), not a single resistor to 2 parallels.
The reason behind selecting the parallel string is due to its orientation. I would use series if I could.
That's what you already have -- two strings of LEDs (with their current-limiting resistor) in parallel.

What is it you are trying to achieve that the current circuit does not accomplish?

#### eetech00

Joined Jun 8, 2013
3,653
I thought they were parallels at best. But its neither? How do I make them parallel then?

I want 2 strings of LEDs in parallel and a power LED power on/off(not shown), not a single resistor to 2 parallels.
The reason behind selecting the parallel string is due to its orientation. I would use series if I could.
Hi

What is implied by the phrase "a string of leds" is a number of LEDs connected in a series fashion so that an equal amount of current flows thru the whole string. There are two "strings" of LED's on the drawing in post #1 and each string has it own dedicated current limiting resistor. Each string would require a separate amount of current from the voltage source.
Since both strings are connected to the same voltage source, technically, the strings, together with the voltage source, are all connected together in a parallel fashion.

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Thread Starter

#### clangray

Joined Nov 4, 2018
261
Each string would require a separate amount of current from the voltage source.
I think this is true in schematic in post #1 - right?

#### eetech00

Joined Jun 8, 2013
3,653
I think this is true in schematic in post #1 - right?
Yes. Each string consists of two LED's and one resistor. Each will draw current separate from the other string.
You can calculate the total current for one string using ohms law. Then, since the strings are identical and the parts are identical, just multiply the total current for one string by 2 to get the total current draw from the voltage source. If the LED's and resistors are different types and values, then the calculation is a little more complicated.

Thread Starter

#### clangray

Joined Nov 4, 2018
261
Yes. Each string consists of two LED's and one resistor. Each will draw current separate from the other string.
You can calculate the total current for one string using ohms law. Then, since the strings are identical and the parts are identical, just multiply the total current for one string by 2 to get the total current draw from the voltage source. If the LED's and resistors are different types and values, then the calculation is a little more complicated.
If I wanted to check the power of the circuit, say 1/4, 1/2 or higher watt resistor, do I check the power of the entire circuit or the individual components (using for ex. P =I^2 x R)?

#### BobTPH

Joined Jun 5, 2013
8,104
If you want to know what wattage resistor to use, you calculate the power dissipated in that resistor only, the rest if yhe circuit dies not matter.

#### eetech00

Joined Jun 8, 2013
3,653
If I wanted to check the power of the circuit, say 1/4, 1/2 or higher watt resistor, do I check the power of the entire circuit or the individual components (using for ex. P =I^2 x R)?
For a serial LED string you only need to calculate the power dissipation of the resistor.

For example, to calculate the Resistor value (assume the LEDs are the same type):
Vsup=5V <---supply voltage
Vf=2.0V <---LED Forward voltage drop
If=0.02A <---LED Forward current (20mA)
nLeds=2 <---Number of LEDs in string

Vd=nLeds x Vf=4.0v <---total LED voltage drop

Rlimit=(Vsup-Vd) / If = 50 ohms <---current limit Resistor value (or next larger standard size)

You will want to calculate the power dissipation of the resistor to determine its wattage rating.
I=0.020 Amps (or 20mA) <---current thru the string
Rlimit=50 ohms <---current limit resistor
Pd=0.020^2 x 50 = 0.020 Watts (or 20mW)

So... use standard value 0.125 Watt (1/8 watt) or higher resistor power rating. Tolerance=2% or less.

The LED circuit would look like this:

Edit: Changed "R" in calculation to "Rlimit" to avoid confusion.

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Thread Starter

#### clangray

Joined Nov 4, 2018
261
R=50 ohms <---current limit resistor
How did you arrive at 50 ohms? Is this an arrived value or supplied value?

Thread Starter

#### clangray

Joined Nov 4, 2018
261
In order to be in parallel, they have to be connected such that the same voltage that appears across one must appear across the other. Not the same value of voltage, but the actual same voltage (sometimes referred to as the same symbolic voltage).

That's what you already have -- two strings of LEDs (with their current-limiting resistor) in parallel.

What is it you are trying to achieve that the current circuit does not accomplish?
I wanted to put this circuit out there and make sure I was constructing parallel strings correctly.

#### BobTPH

Joined Jun 5, 2013
8,104
Yes, the circuit is correct. What has caused confusion is the wording of your questions.

#### eetech00

Joined Jun 8, 2013
3,653
How did you arrive at 50 ohms? Is this an arrived value or supplied value?
Each component of the series string (the resistor and each of the two LEDs) will have a voltage drop associated with it.
The resistor value is calculated based on the required voltage drop across it to satisfy Kirchhoff's voltage law.
You can read about KVL here:

https://www.allaboutcircuits.com/textbook/direct-current/chpt-6/kirchhoffs-voltage-law-kvl/

I've shown how the resistor value is calculated in post #15.

I have no idea of the type of LED, and Vf, If, you used so I used common LED values for the example calculations.

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#### eetech00

Joined Jun 8, 2013
3,653
How did you arrive at 50 ohms? Is this an arrived value or supplied value?
I've changed "R" to "Rlimit" for consistancy. Maybe that confused you a little.